LINEAR OPERATORS OF BOUNDED ON SEMI-UNITARY SPACES

Let U be a finite-dimensional vector space over the field C. The vector space U is a semi-unitary space with semi-inner product [·, ·] if and only if U is an inner product space with inner product ?·, ·? and there exists a A linear operator on U which is positive semi-definite such that for every...

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Bibliographic Details
Main Author: Yusuf Pangestu, Qori
Format: Theses
Language:Indonesia
Online Access:https://digilib.itb.ac.id/gdl/view/73492
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Institution: Institut Teknologi Bandung
Language: Indonesia
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Summary:Let U be a finite-dimensional vector space over the field C. The vector space U is a semi-unitary space with semi-inner product [·, ·] if and only if U is an inner product space with inner product ?·, ·? and there exists a A linear operator on U which is positive semi-definite such that for every x, y ? U, it holds [x, y] = ?Ax, y?. In this thesis, a bounded linear operator on semi-unitary spaces will be studied where the sufficient and necessary conditions of the linear operator A is bounded (i.e. ?Au? ? c?u? for a positive number c ? R and for all u ? U) is the isotropic subspace U0 is A-invariant. Suppose that T is a linear operator where U is a complex semi-inner product space, the adjoint linear operator of T is defined as the linear operator S which satisfies the property that for every x, y ? U, [Tx, y] = [x, Sy] holds. However, the adjoin operator of T does not necessarily exist. Even if there is an adjoin operator, the adjoin operator is not necessarily unique. Therefore, the sufficient and necessary condition for the existence of an adjoin operator is the linear operator being bounded. Suppose the set of all bounded linear operators B(U) that form a nonempty set, so the set of bounded operators B(U) forms a vector space. Since the set B(U) forms a vector space, This vector space has subspaces, one example of a subspace of B(U) is S = {f : U ? U | f a bounded linear operators of U with f(U) ? U0}.